Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $z \neq 0$. $n = \dfrac{-2z - 18}{-3z^3 - 18z^2 + 81z} \times \dfrac{z^3 - 4z^2 + 3z}{z + 9} $
First factor out any common factors. $n = \dfrac{-2(z + 9)}{-3z(z^2 + 6z - 27)} \times \dfrac{z(z^2 - 4z + 3)}{z + 9} $ Then factor the quadratic expressions. $n = \dfrac {-2(z + 9)} {-3z(z - 3)(z + 9)} \times \dfrac {z(z - 3)(z - 1)} {z + 9} $ Then multiply the two numerators and multiply the two denominators. $n = \dfrac {-2(z + 9) \times z(z - 3)(z - 1) } { -3z(z - 3)(z + 9) \times (z + 9)} $ $n = \dfrac {-2z(z - 3)(z - 1)(z + 9)} {-3z(z - 3)(z + 9)(z + 9)} $ Notice that $(z - 3)$ and $(z + 9)$ appear in both the numerator and denominator so we can cancel them. $n = \dfrac {-2z\cancel{(z - 3)}(z - 1)(z + 9)} {-3z\cancel{(z - 3)}(z + 9)(z + 9)} $ We are dividing by $z - 3$ , so $z - 3 \neq 0$ Therefore, $z \neq 3$ $n = \dfrac {-2z\cancel{(z - 3)}(z - 1)\cancel{(z + 9)}} {-3z\cancel{(z - 3)}\cancel{(z + 9)}(z + 9)} $ We are dividing by $z + 9$ , so $z + 9 \neq 0$ Therefore, $z \neq -9$ $n = \dfrac {-2z(z - 1)} {-3z(z + 9)} $ $ n = \dfrac{2(z - 1)}{3(z + 9)}; z \neq 3; z \neq -9 $